602 Proceedings of the Royal Society of Edinburgh. [Sess. 
Adding the last two equations, we get the required result, which proves 
the lemma. 
§ 13. Theorem. — If in the interval (q, go ) the function f(u) can be 
expressed as the product of two functions, one of which, g(u), is monotone 
decreasing with zero as limit at infinity, and the other, h(u), is any periodic 
function whose square is summable in every finite interval, then 
dvf° q e~ kvn f(u) cos uv du = 0, 
provided f ~~du exists. 
For, by the lemma of § 12, 
lo dv\qe~ kv2t f(u) cos uv du = J Q duj^ e~ kv2t /(u) cos uv dv 
f” ,, x sinB'w 7 
= flu) du, 
J Q U 
using the Second Theorem of the Mean, B' lying between 0 and B. 
the integral on the left is numerically 
Hence 
z rm du , 
J Q U 
which, by the hypothesis that 
Q recedes to infinity. 
Moreover, by a similar argument, 
j du exists, has zero as unique limit when 
7 Tcb^t rt\u 
cos uv du — e K0 1 / • A_ 
J Q U 
du. 
which vanishes when b increases without limit. 
Therefore J* dv\”e~ kv2t f(u) cos uv du exists, and, by the above 
j dvj e~ kv2t f{u) cos uvdu | <j du . . (1) 
But, in any finite interval ( q , Q), h(u) is summable and g(u) is bounded ; 
therefore f(u) is summable. Hence, by § 2, 
dv\^e~ kv2t f(u) cos uv du = 0 .... (2) 
Adding (1) and (2) 
I ht [ dv f e kv2t f(u) cos uv du I L f du . 
I t=oJ 0 J q \ J Q U 
But the left-hand side is independent of Q, and the right-hand side 
vanishes when Q moves off to infinity ; hence the required result follows. 
